Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $q \neq 0$. $y = \dfrac{5q + 15}{-q^3 - 13q^2 - 30q} \times \dfrac{q^3 + 8q^2 - 20q}{q - 8} $
Answer: First factor out any common factors. $y = \dfrac{5(q + 3)}{-q(q^2 + 13q + 30)} \times \dfrac{q(q^2 + 8q - 20)}{q - 8} $ Then factor the quadratic expressions. $y = \dfrac {5(q + 3)} {-q(q + 10)(q + 3)} \times \dfrac {q(q + 10)(q - 2)} {q - 8} $ Then multiply the two numerators and multiply the two denominators. $y = \dfrac {5(q + 3) \times q(q + 10)(q - 2) } { -q(q + 10)(q + 3) \times (q - 8)} $ $y = \dfrac {5q(q + 10)(q - 2)(q + 3)} {-q(q + 10)(q + 3)(q - 8)} $ Notice that $(q + 10)$ and $(q + 3)$ appear in both the numerator and denominator so we can cancel them. $y = \dfrac {5q\cancel{(q + 10)}(q - 2)(q + 3)} {-q\cancel{(q + 10)}(q + 3)(q - 8)} $ We are dividing by $q + 10$ , so $q + 10 \neq 0$ Therefore, $q \neq -10$ $y = \dfrac {5q\cancel{(q + 10)}(q - 2)\cancel{(q + 3)}} {-q\cancel{(q + 10)}\cancel{(q + 3)}(q - 8)} $ We are dividing by $q + 3$ , so $q + 3 \neq 0$ Therefore, $q \neq -3$ $y = \dfrac {5q(q - 2)} {-q(q - 8)} $ $ y = \dfrac{-5(q - 2)}{q - 8}; q \neq -10; q \neq -3 $